How can you show the results from substituting $x=a\cos{\theta}$ and $x=a\sin{\theta}$ are the same when finding $\int{\frac{16}{\sqrt{9-x^2}}}$?

I am learning trig substitution right now. I know from seeing this problem I should substitute x for $a\sin{\theta}$. After working this out:

$$\int{\frac{16}{\sqrt{9-x^2}}}$$ $$x=3\sin{\theta}$$ $$dx=3\cos{\theta}d\theta$$ $$=\int{\frac{16}{\sqrt{9-x^2}}}=\int{\frac{48\cos{\theta}d\theta}{\sqrt{9-9\sin^2}{\theta}}}$$ $$=\int{\frac{48\cos{\theta}d\theta}{3\sqrt{1-\sin^2}{\theta}}}$$$$=16\int{\frac{\cos{\theta}d\theta}{\sqrt{\cos^2}{\theta}}}$$ $$=16\int{\frac{\cos{\theta}d\theta}{\cos{\theta}}}=16\int{d\theta}$$ $$=16\theta=16\sin^{-1}\left({\frac{x}{3}}\right)+C_1$$

However, substituting $x=a\cos{\theta}$ yields a different result.

$$\int{\frac{16}{\sqrt{9-x^2}}}$$ $$x=3\cos{\theta}$$ $$dx=-3\sin{\theta}d\theta$$ $$=\int{\frac{16}{\sqrt{9-x^2}}}=\int{\frac{-48\sin{\theta}d\theta}{\sqrt{9-9\cos^2}{\theta}}}$$ $$=\int{\frac{-48\sin{\theta}d\theta}{3\sqrt{1-\cos^2}{\theta}}}$$$$=-16\int{\frac{\sin{\theta}d\theta}{\sqrt{\sin^2}{\theta}}}$$ $$=-16\int{\frac{\sin{\theta}d\theta}{\sin{\theta}}}=-16\int{d\theta}$$ $$=-16\theta=-16\cos^{-1}\left({\frac{x}{3}}\right)+C_2$$


Unless I did something incorrectly or there is something that prohibits different types of substitutions, how can this be true? I have tried to think it over how $-\cos^{-1}{x}$ can equal $\sin^{-1}{x}$.

I thought about it being a negative angle, perhaps making the opposite leg of the triangle negative. This however leads to a contradiction that says $-\cos^{-1}{x}=\sin^{-1}{x}$ but $\cos^{-1}{x}\neq-\sin^{-1}{x}$.

"positive angle"

"possibly a negative angle"

Any thoughts, answers, or corrections are appreciated.

  • 1
    $\begingroup$ There comes "+ arbitrary constant" in both of these indefinite integrals... $\endgroup$
    – Bob Dobbs
    Nov 2, 2022 at 4:06
  • 1
    $\begingroup$ Thank you! I forgot. $\endgroup$ Nov 2, 2022 at 4:09
  • $\begingroup$ The 2 "C" Values are not the same ! The last line is "Something=Something+C" where this C is not the same ! Try using "C1" & "C2" & "C3" to make that rigorous ! $\endgroup$
    – Prem
    Nov 2, 2022 at 4:15
  • $\begingroup$ I'll do that. Thanks. $\endgroup$ Nov 2, 2022 at 4:46
  • 1
    $\begingroup$ Ok , when you changed it to "Something+C1=Something+C2" , you got "Something=Something+(C2-C1)=Something+C3" & this is entirely true in your Case. There is no contradiction : $\sin^{-1}(x)$ & $-\cos^{-1}(x)$ are same except for a Constant. Plot these two to check that Constant ! $\endgroup$
    – Prem
    Nov 2, 2022 at 14:45

2 Answers 2


If you do a simpler example of $\int \frac{1}{\sqrt{1 - x^2}} dx$, you will get $\sin^{-1}(x) + C$ and $-\cos^{-1}(x) + C$ respectively. If you draw the diagrams, you will see that $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$, since it's the two angles that's not a right angle. Therefore, you see that $\sin^{-1}(x)$ and $-\cos^{-1}(x)$ differ by a constant $\frac{\pi}{2}$.

  • $\begingroup$ $+C$ doesn't mean a specific constant, it means a constant. It's like $\int f'(x) dx = f(x) + C_1 = \left(f(x) + 1\right) + C_2$, they look different but they're the same. $\endgroup$
    – Gareth Ma
    Nov 2, 2022 at 5:36

Quite simply $$-(\cos^{-1}x)'=(\sin^{-1}x)'$$ so they differ by a constant. And that constant is easily determined to be $$\pi/2$$.

Everything checks out in terms of symmetry on the unit circle.

Just remember that when you compute an indefinite integral, the answer only holds up to a constant.


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